16cos^5A+2cosA-cos3A-cos5A=16cos^3A

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16cos^5A+2cosA-cos3A-cos5A=16cos^3A 0:00:12

16cos^5A+2cosA-cos3A-cos5A=16cos^3A

Prove that: cos 0:07:48

Prove that: cos 5A = 16cos^5 A - 20cos^2 A + 5cos A

Prove that: `cos5A=16cos^5A-20cos^3A+5cosA` 0:04:53

Prove that: `cos5A=16cos^5A-20cos^3A+5cosA`

Cos^(3)A. Sin^(2)A = 0:03:49

Cos^(3)A. Sin^(2)A = 1/16 (2cosA - cos3A -cos5A)

Prove that :- 0:07:50

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cos(2A)cos(3A)−cos(2A)cos(7A)sin(4A)sin(3A)−sin(2A)sin(5A)=sin(7A)+sin(3A)sinA 0:00:28

cos(2A)cos(3A)−cos(2A)cos(7A)sin(4A)sin(3A)−sin(2A)sin(5A)=sin(7A)+sin(3A)sinA

`2costheta-cos 3 theta-cos5 0:02:10

`2costheta-cos 3 theta-cos5 theta-16cos^3 theta sin^2 theta=`

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Some examples Type:1 0:06:19

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TRIGONOMETERY PROVE THAT 1 + cos 2A + cos 4A + cos 6A = 4 cos A cos 2A cos 3A.

[Nepali] Evaluate Cos5A 0:07:12

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Prove that cos3Acos(3A)+sin3A(sin3A)=cos32A 0:09:25

Prove that cos3Acos(3A)+sin3A(sin3A)=cos32A

How to solve 0:02:24

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Prove that: cos2Pi/15 . cos4Pi/15 . cos8Pi/15 . cos16Pi/15 = 1/16

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(cos^3x.sin^2x)=1/16 (2cosx-cos3x-cos5x) I 0:04:04

(cos^3x.sin^2x)=1/16 (2cosx-cos3x-cos5x) I XI I Trigonometry I CBSE I ICSE I NCERT I Tough Question

Prove2cos4A = (2cosA+1)(2cosA-1)(2cos2A-1) 0:05:03

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Prove that: sin⁴ A = 3/8 - 1/2 cos 2A + 1/8 cos 4A

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if `2tanx=3tany` , show that `tan(x-y)=(sin2y)/(5-cos2y)`

Find the value 0:06:38

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